Page 41 - PRE-U STPM MATHEMATICS (T) TERM 1

Mathematics Term 1 STPM Chapter 2 Sequences and Series

(b) Substitute x = –2x and n = –3, we get

(1 – 2x) = 1 + (–3)(–2x) + (–3)(–3 – 1) (–2x) + (–3)(–3 – 1)(–3 – 2) (–2x)
3
2
–3
2! 3!
4
+ (–3)(–3 – 1)(–3 – 2)(–3 – 3) (–2x) + …
4!

2
= 1 + 6x + (–3)(–4) (4x ) + (–3)(–4)(–5) (–8x )
3
2 6
+ (–3)(–4)(–5)(–6) (16x ) + …
4
24
4
2
3
= 1 + 6x + 24x + 80x + 240x + …
2
The expansion is valid if |–2x| , 1, i.e. 2|x| , 1,
1
or |x| , 1 or – , x , 1 .
2 2 2
Example 39

10
10
3
Expand (1 + x) up to and including the term in x . Hence, obtain an approximation for (1.01) and (0.99) .
10
2
3
Solution: (1 + x) = 1 + 10x + 10 · 9 x + 10 · 9 · 8 x + …
10
1 · 2 1 · 2 · 3
3
2
= 1 + 10x + 45x + 120x + …
Substitute x = 0.01 into the expansion,
3
2
10
(1 + 0.01) = 1 + 10(0.01) + 45(0.01) + 120(0.01) + …
= 1 + 0.1 + 0.0045 + 0.000120 + …
= 1.10462
= 1.1046 (correct to 4 decimal places)

Substitute x = –0.01 into the expansion,

3
2

10

(1 – 0.01) = 1 + 10(–0.01) + 45(–0.01) + 120(–0.01) + …
= 1 – 0.1 + 0.0045 – 0.000120 + …
= 0.90438
= 0.9044 (correct to 4 decimal places)
10
10
Note: By using the calculator, we find that (1.01) = 1.1046 and (0.99) = 0.9044 (both correct to 4 decimal
places). This shows that the accuracy of the approximation does not differ much even if we were to
10
expand (1 + x) up to x or more. Hence, usually three or four terms should be sufficient for a good
5
approximation.

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02 STPM Math T T1.indd 128 3/28/18 4:21 PM